What is the integral from 0 to 5 of x/(sqrt(9-x)) ?

The integral from 0 to 5 of x/(sqrt(9-x)) is 16/3

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integral from 0 to 5 of x/(sqrt(9-x))

\int_{0}^{5} \frac{x}{9 - x} d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{5} \frac{x}{9 - x} d x

Apply u-substitution

: \int_{3}^{2} - 2 (- u^{2} + 9) d u

= \int_{3}^{2} - 2 (- u^{2} + 9) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{3} - 2 (- u^{2} + 9) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- 2 \cdot \int_{2}^{3} - u^{2} + 9 d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- 2 (- \int_{2}^{3} u^{2} d u + \int_{2}^{3} 9 d u))

\int_{2}^{3} u^{2} d u = \frac{19}{3}

\int_{2}^{3} 9 d u = 9

= - (- 2 (- \frac{19}{3} + 9))

Simplify

- (- 2 (- \frac{19}{3} + 9)) : \frac{16}{3}

= \frac{16}{3}

y = 4 - x^{2}, y = - x + 2

\int \frac{x - 1}{2 x + 4} d x

\frac{d}{d x} (- \frac{2}{x ^{3}} + \frac{3}{x ^{2}} - 4 x)

\int cos^{4} (4 x) d x

\frac{\partial}{\partial x} (- 2 e^{- 2 x})

What is the integral from 0 to 5 of x/(sqrt(9-x)) ?
The integral from 0 to 5 of x/(sqrt(9-x)) is 16/3